Chapter 4, Problem 63RE

(a)

To determine

To find:

The beam of maximal cross sectional area is square.

Answer to Problem 63RE

The beam of maximal cross sectional area is square.

  $A=x\sqrt{4r^2-x^2}$

Explanation of Solution

Given:

The radius of the cylinder

  $r=10\text{inches}$

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let $x$ be depth of the rectangle and $y$ be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length $2r$ .

By using Pythagoras theorem

  $\begin{array}{l}{x}^2+y^2={\left(2r\right)}^2\\ y^2={\left(2r\right)}^2-x^2\\ y=\sqrt{4r^2-x^2}\end{array}$

The area of the rectangule $A=xy\mathrm{................................}\left(1\right)$

Now after substituting $\sqrt{4r^2-x^2}$ for $y$ in equation (1)

  $A=x\sqrt{4r^2-x^2}$

(b)

To determine

To find:

The dimensions of the planks that will have maximal cross-sectional area.

Answer to Problem 63RE

The dimensions of the planks that will have maximal cross-sectional area is $x=2r$ .

Explanation of Solution

Given:

The radius of the cylinder

  $r=10\text{inches}$

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let $x$ be depth of the rectangle and $y$ be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length $2r$ .

By using Pythagoras theorem

  $\begin{array}{l}{x}^2+y^2={\left(2r\right)}^2\\ y^2={\left(2r\right)}^2-x^2\\ y=\sqrt{4r^2-x^2}\end{array}$

The area of the rectangule $A=xy\mathrm{................................}\left(1\right)$

Now after substituting $\sqrt{4r^2-x^2}$ for $y$ in equation (1)

  $A=x\sqrt{4r^2-x^2}\mathrm{...............................}\left(2\right)$

Differentiating with respect to $x$

  $\begin{array}{l}\frac{dA}{dx}=\frac{d}{dx}x\sqrt{4r^2-x^2}\\ \frac{dA}{dx}=x\frac{d}{dx}\sqrt{4r^2-x^2}+\sqrt{4r^2-x^2}\frac{dx}{dx}\\ \frac{dA}{dx}=\sqrt{4r^2-x^2}-\frac{2x^2}{\sqrt{4r^2-x^2}}\end{array}$

Using the zero product property of the equation

  $\begin{array}{l}0=\sqrt{4r^2-x^2}-\frac{2x^2}{\sqrt{4r^2-x^2}}\\ 0=4r^2-x^2-2x^2\\ 0=4r^2-x^2\\ x=\pm 2r\\ x=2r\end{array}$

(c)

To determine

To find:

The dimensions of the strongest beam that cut from the cylindrical log.

Answer to Problem 63RE

The dimensions of the strongest beam that cut from the cylindrical log is $x\sqrt{4r^2-x^2}=y$

Explanation of Solution

Given:

The radius of the cylinder

  $r=10\text{inches}$

Concept used:

If the quantities $x$ and $y$ are related by an equation

  $y\alpha x$

For some constant $k\ne 0$

That is y varies inversely proportional as $x$ or y is directly proportional to $x$

The constant $k$ is called constant proportionality

Calculation:

Let $x$ be depth of the rectangle and $y$ be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length $2r$ .

By using Pythagoras theorem

  $\begin{array}{l}{x}^2+y^2={\left(2r\right)}^2\\ y^2={\left(2r\right)}^2-x^2\\ y=\sqrt{4r^2-x^2}\end{array}$

The area of the rectangule $A=xy\mathrm{................................}\left(1\right)$

Now after substituting $\sqrt{4r^2-x^2}$ for $y$ in equation (1)

  $A=x\sqrt{4r^2-x^2}$

If the quantities $x$ and $y$ are related by an equation

  $\begin{array}{l}y\alpha x\\ y=kx\\ x\sqrt{4r^2-x^2}=y\end{array}$

For some constant $k\ne 0$